The sum \sum _ { k = 1 } ^ { \infty } \frac { 1 } { k ^ { 3 } + 6 k ^ { 2 } + 8 k } can be expressed as \frac { a } { b }, where a and b are relatively prime positive integers. What is a+b?
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\sum_{k=1}^{\infty} \frac{k}{k(k+2)(k+4)} = \sum_{k=1}^{\infty} \frac{1}{(k+2)(k+4)}
= \frac{1}{2} \sum_{k=1}^{\infty} \frac{(k+4)-(k+2)}{(k+2)(k+4)} = \frac{1}{2} \sum_{k=1}^{\infty} \left( \frac{1}{k+2} - \frac{1}{k+4} \right)
= \frac{1}{2} \cdot \left( \frac{1}{3} + \frac{1}{4} \right)
Similarly:
\sum_{k=1}^{\infty} \frac{k+4}{k(k+2)(k+4)} = \sum_{k=1}^{\infty} \frac{1}{k(k+2)} = \frac{1}{2} \cdot \left(1 + \frac{1}{2}\right)
Therefore:
\sum_{k=1}^{\infty} \frac{1}{k(k+2)(k+4)} = \frac{1}{4} \cdot \left( \sum_{k=1}^{\infty} \frac{k+4}{k(k+2)(k+4)} - \sum_{k=1}^{\infty} \frac{k}{k(k+2)(k+4)} \right)
= \frac{1}{8} \cdot \left( 1 + \frac{1}{2} - \frac{1}{3} - \frac{1}{4} \right) = \frac{11}{96}
